Ladders 1 and 2, denoted L1 and L2, respectively, will rest along two
walls (taken to be perpendicular to the ground), and they will
intersect at a point O = (a,s), a height s from the ground. Find the
largest s such that this is possible. Then find the width of the
alley, w = a+b, in terms of L1, L2, and s. This diagram is not to
scale.

Now, (**) defines a fourth degree polynomial in y. It can be written in the
form (by simply expanding (**))

(***) y^4 - 2s_y^3 - L_y^2 + 2sL_y - Ls^2 = 0

L1 and L2 are given, and so L is a constant. How large can s be? Given L,
the value s=k is possible if and only if there exists a real solution, y',
to (***), such that 2k <= y' < L2. Now that s has been chosen, L and s are
constants, and (***) gives the desired value of y. (Make sure to choose the
value satisfying 2s <= y' < L2. If the value of s is "admissible" (i.e.,
feasible), then there will exist exactly one such solution.)
Now, w = sqrt(L2^2 - y^2), so this concludes the solution.

L1 = 11, L2 = 13, s = 4. L = 13^2-11^2 = 48, so (***) becomes

y^4 - 8_y^3 - 48_y^2 + 384_y - 768 = 0

Numerically find root y = 9.70940555, which yields w = 8.644504.

Owner: "70.187.170.60" Last edited on August 2, 2005 6:10 pm by "70.187.170.60"